25. Stack of Blocks on Inclined Plane A stack of3 identical blocks is placed on

Question

25. Stack of Blocks on Inclined Plane A stack of3 identical blocks is placed on an inclined plane as shown in the figure. The plane is at an angle from the horizontal. Each block has a mass of 1.5 kg and the coefficient of static friction between all surfaces is ,-0.8. For convenience, the blocks are labeled 1,2 and 3 as indicated. What is the net force on block 27 Why? (b) Find the total force which block 3 exerts on block 2. Give the magnitude and direction of the force. Explain your answer. (c) Find the frictional force which block 1 exerts on block 2. Give the magnitude and direction of the force.

Explanation / Answer

Forces acting on block -2

1. Its weight mg - vertically down

2. Normal reaction of block2 - mgCos(12)

3. Frictional force at the lower face btween 1 and 2

4. Normal force by block-3

5. frictional force on the top surface 2 and 3

resolving the forces vertically and horizontally

N12 = mgCos(12)

N23 = mgCos(12)

Frictional force between 1 and2

F12 = 0.8(N12 + N23) = 0.8*2*1.5*9.8Cos(12) = 23 N

Frictional force between 2 and 3

F23 = 0.8*9.8*1.5*Cos(12) = 11.5 N

weight parallel to the incline = mgSin(12) = 1.5*9.8*sin(12) = 3.06 N

The two frictional force oppose the motion of the block and act up the incline whereas the weight component act down the incline

The net force on the block = 3.06 - 11.5-23

The frictional force is the limiting friction is limited to the wieght of the block and hence net force =0

b) Force by block-3 on block2

weight component mg - vertically down

Frictional fiorce 0.8N

N = mgCos(12) = 14.38 N

Frictional force = 0.8*14.38 = 11.5 N , up the incline

c) Normal reaction from Block-1     = 2mgCos(12)

Frictional force F12 = 0.8N = 0.8*2*1.5*9.8Cos(12) = 23 N

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